1(2) i guess is solved here http://www.targetiit.com/iit-jee-forum/posts/aieee-ques-15797.html
by gallardo
(!) f(x) = Lt n-->∞ [ x/(1+(2sinx)2n) ]
then f is discontinuous at
(A) pi
(B) pi/3
(C) pi/4
(D) pi/6
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6 Answers
1(2) The fraction exceeding its pth power by the greatest no. possible where p≥2 is
(1) (1/p)1/(p-1)
(2) (1/p)p-1
(3) p1/(p-1)
(4) pp-1
1357The first one will have discontinuity when sin x becomes 1/2 or -1/2
can you figure out why?
62The second one is x-xp has to be maximized.. given that x lies in 0 and 1
so the derivative is zero... and double derivative is -ve is obvious
1-pxp-1
So x = (1/p)1/(p-1)
11@Manish Siir for first one i coudnt figure out y ?? cud u explain that plzz
@Nishant Sir : Thanks for the 2 nd one
got it!!
62maya, that post was by me.. by mistake posted by manish''s ID...
see the thing there is imple there is (a)^n where n tends to infinty..
We have a discontinuity when a equals to 1, because if a is less than 1 then the limit will give zero, at a=1, the limit will be 1 and when a greater than 1, the limit will be + infinity..
a=(2 sin x)
More over we have an even power..
Hence the conclusion